If $\;$ $x + \dfrac{1}{x} = 2 \cos \theta$, $\;$ prove that $\;$ $x^n + \dfrac{1}{x^n} = 2 \cos \left(n \theta\right)$ $\;$ and $\;$ $x^n - \dfrac{1}{x^n} = 2 i \sin \left(n \theta\right)$
Let $x = \cos \theta + i \sin \theta$
$\begin{aligned}
Then, \; \dfrac{1}{x} & = \dfrac{1}{\cos \theta + i \sin \theta} \\\\
& = \dfrac{\cos \theta - i \sin \theta}{\left(\cos \theta + i \sin \theta\right) \left(\cos \theta - i \sin \theta\right)} \\\\
& = \dfrac{\cos \theta - i \sin \theta}{\cos^2 \theta - i^2 \sin^2 \theta} \\\\
& = \cos \theta - i \sin \theta
\end{aligned}$
so that $\;$ $x + \dfrac{1}{x} = 2 \cos \theta$
Now,
$x^n = \left(\cos \theta + i \sin \theta\right)^n = \cos \left(n \theta\right) + i \sin \left(n \theta\right)$
$\dfrac{1}{x^n} = \left(\cos \theta - i \sin \theta\right)^n = \cos \left(n \theta\right) - i \sin \left(n \theta\right)$
$\therefore \; x^n + \dfrac{1}{x^n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) + \cos \left(n \theta\right) - i \sin \left(n \theta\right) = 2 \cos \left(n \theta\right)$
$x^n - \dfrac{1}{x^n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) - \cos \left(n \theta\right) + i \sin \left(n \theta\right) = 2 i \sin \left(n \theta\right)$
Hence proved.